If $x$ is numerically so small that $x^2$ and higher powers of $x$ can be neglected, then $\left(1+\frac{2x}{3}\right)^{3/2} \cdot (32+5x)^{-1/5}$ is approximately equal to
- A$\frac{32+31x}{64}$
- B$\frac{31+32x}{64}$
- C$\frac{31-32x}{64}$
- D$\frac{1-2x}{64}$
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Similar Questions
$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :
DifficultAP EAMCET 2006
View SolutionIf $-\frac{2}{3} < x < \frac{2}{3}$,then the value of the $5^{\text{th}}$ term in the expansion of $\frac{1}{\sqrt[3]{2-3x}}$ when $x=\frac{1}{2}$ is
Assertion $(A)$: If $|x| < 1$,then $\sum_{n=0}^{\infty}(-1)^n x^{n+1} = \frac{x}{x+1}$.
Reason $(R)$: If $|x| < 1$,then $(1+x)^{-1} = 1-x+x^2-x^3+\dots$.
Which one of the following is true?
Reason $(R)$: If $|x| < 1$,then $(1+x)^{-1} = 1-x+x^2-x^3+\dots$.
Which one of the following is true?
If $x = \frac{2 \cdot 5}{(2!) 3} + \frac{2 \cdot 5 \cdot 7}{(3!) 3^2} + \frac{2 \cdot 5 \cdot 7 \cdot 9}{(4!) 3^3} + \dots$,then $x^2 + 8x + 8 = $
If $x=\frac{2}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3+\ldots$,then $x+\frac{1}{x}=$
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